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I can't wrap my head around this probability problem. I think I have to use the Binomial Theorem to solve this problem, but I can't figure out how, now that I am not given n or p. The problem states:

A large group of 16 to 24-year-olds were asked if they had consumed alcohol within the last year. Of the people asked 43.1% were men and 56.9% were women. 8.1% of the asked men answered that they had not consumed alcohol within the last year and 10.2% women answered that they had not consumed alcohol within the last year.

Question 1: Compute the probability that a randomly selected 16 to 24-year-old has not consumed alcohol within the last year.

If any of you guys might have a hint of what to do it would be much appreciated.

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  • $\begingroup$ It appears that a word is missing. "$8.1\%$ of the " what? (I suspect "men", but please verify.) $\endgroup$
    – Graham Kemp
    Oct 2, 2016 at 13:10
  • $\begingroup$ PS: This has naught to do with the Binomial Theorem. Nor is it a Binomial Distribution for that matter. $\endgroup$
    – Graham Kemp
    Oct 2, 2016 at 13:13
  • $\begingroup$ the question cannot be the correction of the sample based on an estimation of the population, because the ratio men/women is not 1, the data misses. It cannot be a matter of probability of a new sample because too much datas miss too. Is the question complete ? $\endgroup$
    – user354674
    Oct 2, 2016 at 13:38

2 Answers 2

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Hint: Use the Law of Total Probability

$$\mathsf P(A)=\mathsf P(A\mid B)\mathsf P(B)+\mathsf P(A\mid B^\complement)\mathsf P(B^\complement)$$

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  • $\begingroup$ Sorry, but I am not sure I follow. Can you show me how to do this numerically? $\endgroup$
    – DBE7
    Oct 2, 2016 at 14:20
  • $\begingroup$ @DBE7 Let $A$ be the event that the person has not consumed alcohol. Let $B$ be the event that the person is a man, and $B^\complement$ that they are a woman. Evaluate the required probabilities in the RHS from the data given, and then calculate the LHS. $\endgroup$
    – Graham Kemp
    Oct 2, 2016 at 18:25
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    $\begingroup$ 0.0929 (=0.081*0.431+0.102*0.569), does this seem right? @GrahamKemp $\endgroup$
    – DBE7
    Oct 2, 2016 at 18:33
  • $\begingroup$ @DBE7 Yes, indeed. $\endgroup$
    – Graham Kemp
    Oct 2, 2016 at 18:39
  • $\begingroup$ That's perfect. Thank you, sir. The challenge I often find difficult is to distinguish between two events' intersect and the probability of one event given another ... $\endgroup$
    – DBE7
    Oct 2, 2016 at 18:41
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Hint: Suppose you were not given the percentages of men and women asked, but can assume the relevant population contains 50% men and 50% women.

I'm assuming here that the large group is just a sample, not the total population of interest.

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